Differential Equations: Second Order Equations

In summary, the problem involves finding a particular solution for the given differentiable equation y''+5y'+6y=4-t^2. The attempt at a solution involved using the same form of a polynomial of degree 1 for the right-hand side, but the initial guess of y=a(t^2)+b was incomplete. The correct guess is y=at^2+bt+c, as it allows for the cancellation of terms on the left side.
  • #1
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Homework Statement



find a particular solutions for the given differentiable equation.

Homework Equations



y''+5y'+6y=4-t^2

The Attempt at a Solution



Because the right-hand side is a polynomial of degree 1, so I want to have a particular solution of the same form. It's like y=a(t^2)+b, but if so, I kinda stuck in the middle of calculation. 6a=-1 (this is fine) and 2a+10at+6b=4 (I want to take out t). i need to use this technique. Please help me find what's wrong here.

Thank you so much!
 
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  • #2
it seems that your original guess at the particular solution is a bit lacking. your choice for y(t) is missing something... can you guess what it is missing?
 
  • #3
yea, I realized that, but I don't kno what's missing...that's a problem.
 
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  • #4
You should guess y=at2+bt+c, and I'll even tell you why.

Let y=at2+b

If you plug that into y''+5y'+6y, you get:

2a +5(2at) +6(at2+b) = 4-t2

Now you nothing with which to cancel out that 10at in the middle of the left side.
 
  • #5
I already tried y=at^2+bt+c, but I still didn't have a way t cancel out t...
 
  • #6
Oh, wait! I don't have to cancel out t...!lol I was too stupid...
 

FAQ: Differential Equations: Second Order Equations

What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a function, its derivatives, and its second derivatives. It can be written in the form of y'' = f(x,y,y'), where y' represents the first derivative of y with respect to x, and y'' represents the second derivative.

What is the general solution to a second order differential equation?

The general solution to a second order differential equation is a function that satisfies the equation for all possible values of the independent variable, typically denoted as x. It contains two arbitrary constants, which can be determined by applying initial or boundary conditions.

What are initial and boundary conditions?

Initial conditions are values of the unknown function and its derivatives at a specific point, usually the starting point of the problem. Boundary conditions, on the other hand, are values of the unknown function and its derivatives at the endpoints or boundaries of the problem domain. These conditions are necessary to find the specific solution to a differential equation.

What are the methods for solving second order differential equations?

There are several methods for solving second order differential equations, including separation of variables, substitution, variation of parameters, and series solutions. Each method has its own advantages and is suitable for different types of problems.

What are applications of second order differential equations?

Second order differential equations have various applications in physics, engineering, economics, and many other fields. They are commonly used to describe the motion of objects, such as the swinging of a pendulum or the vibration of a spring. They are also used in modeling population growth, electrical circuits, and chemical reactions.

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